## brinethery 3 years ago Scientists want to place a 2800.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2695.0 m/s in a perfectly circular orbit. Here is some information that may help solve this problem: mass of mars = 6.4191 x 10^23 kg radius of mars = 3.397 x 10^6 m G = 6.67428 x 10^-11 N-m^2/kg^2 What should the speed of the orbit be, if we want the satellite to take 8 times longer to complete one full revolution of its orbit?

1. brinethery

My radius is 5.89876*10^6 m. I'm aware that T = (2pi/sqrt(GM))*r^(3/2) and that circumference = 2pi*r However, when try to solve the problem, I keep coming up with the wrong answer.

2. brinethery

@Algebraic! Would you mind helping on this last one?

3. brinethery

I solved for T, and then multiplied that time by 8. I then found the circumference of the orbit. I divided circumference/8T

4. Algebraic!

I think what you need to do is compare 2*pi*r/V to 16 *pi*r/V

5. Algebraic!

you can use the result from the other problem we did to write r in terms of V I believe...

6. brinethery

I'll be back to check and see what you have to say. :-) Thank you again

7. Algebraic!

$T = \frac{2 \pi r }{ V} = 2 \pi G\frac{ M _{mars} }{ V^3 }$

8. Algebraic!

so the period goes like the inverse cube of velocity....

9. Algebraic!

$\frac{ 8T }{ T } = \frac{ 2695^3 }{ V^3}$ $2695^3 \frac{1}{8} = V^3$ aka half the velocity...

10. brinethery

I was thinking now that if the satellite changes velocity, it falls out of orbit and into a different one. If it takes 8 times as long, does the radius (as well as the orbit) change?

11. brinethery

I'm still working through it...

12. Algebraic!

yes.

13. Algebraic!

It's really more like... you increase the orbital radius, force from gravity is weaker, centripetal force is less, velocity is less... so distance (circumference) is greater and velocity is smaller... that's what that expression I used shows... (I hope)